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hdu 2544 最短路

时间:2018-04-16 22:21:27      阅读:30      评论:0      收藏:0      [点我收藏+]

标签:需要   table   limit   包括   back   所在地   font   i++   sample   

最短路

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 80757    Accepted Submission(s): 34969


Problem Description
在每年的校赛里,所有进入决赛的同学都会获得一件很漂亮的t-shirt。但是每当我们的工作人员把上百件的衣服从商店运回到赛场的时候,却是非常累的!所以现在他们想要寻找最短的从商店到赛场的路线,你可以帮助他们吗?

 
Input
输入包括多组数据。每组数据第一行是两个整数N、M(N<=100,M<=10000),N表示成都的大街上有几个路口,标号为1的路口是商店所在地,标号为N的路口是赛场所在地,M则表示在成都有几条路。N=M=0表示输入结束。接下来M行,每行包括3个整数A,B,C(1<=A,B<=N,1<=C<=1000),表示在路口A与路口B之间有一条路,我们的工作人员需要C分钟的时间走过这条路。
输入保证至少存在1条商店到赛场的路线。
 
Output
对于每组输入,输出一行,表示工作人员从商店走到赛场的最短时间
 
Sample Input
2 1 1 2 3 3 3 1 2 5 2 3 5 3 1 2 0 0
 
Sample Output
3 2
 
思路:最短路水题,练习算法。
  • 不知道什么名字的算法:

 

Run ID Submit Time Judge Status Pro.ID Exe.Time Exe.Memory Code Len. Language Author
24454955 2018-04-16 21:04:36 Accepted 2544 62MS 2024K 808 B C++ zqiangda

 

#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
using namespace std;
struct Node{
    int s, e, w;
    Node(int a, int b, int c) :s(a), e(b), w(c){}
};
int main()
{
    int n, m;
    long len[105];
    vector<Node>vec;
    while (cin >> n >> m, n+m){
        vec.clear();
        for (int i = 0; i < m; i++){
            int a, b, c;
            cin >> a >> b >> c;
            vec.push_back(Node(a, b, c));
        }
        memset(len, 125, sizeof(len));
        len[1] = 0;

        int flag = 1;
        while (flag){
            flag = 0;
            for (int i = 0; i < vec.size(); i++){
                int s = vec[i].s, e = vec[i].e, w = vec[i].w;

                if (len[e]>len[s] + w){
                    len[e] = len[s] + w;
                    flag = 1;
                }
                if (len[s] > len[e] + w){
                    len[s] = len[e] + w;
                    flag = 1;
                }
            }
        }
        cout << len[n] << endl;
    }
    return 0;
}

 

  •  SPFA算法:
Run ID Submit Time Judge Status Pro.ID Exe.Time Exe.Memory Code Len. Language Author
24455250 2018-04-16 21:19:34 Accepted 2544 62MS 2068K 974 B C++ zqiangda
#include<cstdio>
#include<vector>
#include<cstring>
#include<queue>
#include<iostream>
using namespace std;
struct Node{
    int e, w;
    Node(int b, int c) :e(b), w(c){}
};
int n, m;
long len[105], vis[105];
vector<Node>vec[105];
void spfa(int s)
{
    memset(len, 125, sizeof(len));
    memset(vis, 0, sizeof(vis));
    len[s] = 0;
    queue<int>que;
    que.push(s);
    while (!que.empty()){
        int p = que.front(); que.pop();
        vis[p] = 0;
        for (int i = 0; i < vec[p].size(); i++){
            int e = vec[p][i].e, w = vec[p][i].w;
            if (len[e]>len[p] + w){
                len[e] = len[p] + w;
                if (!vis[e]){
                    que.push(e);
                    vis[e] = 1;
                }
            }

        }
    }

}
int main()
{
    
    while (cin >> n >> m, n+m){
        for (int i = 1; i <= n;i++)
            vec[i].clear();
        for (int i = 0; i < m; i++){
            int a, b, c;
            cin >> a >> b >> c;
            vec[a].push_back(Node(b, c));
            vec[b].push_back(Node(a, c));
        }
        
        spfa(1);
        cout << len[n] << endl;
    }
    return 0;
}
  • dijkstra算法:
Run ID Submit Time Judge Status Pro.ID Exe.Time Exe.Memory Code Len. Language Author
24455452 2018-04-16 21:32:54 Accepted 2544 46MS 2064K 1123 B C++ zqiangda
#include<cstdio>
#include<vector>
#include<cstring>
#include<climits>
#include<iostream>
using namespace std;
struct Node{
    int e, w;
    Node(int b, int c) :e(b), w(c){}
};
int n, m;
long len[105], vis[105];
vector<Node>vec[105];
void dijkstra(int s)
{
    memset(len, 125, sizeof(len));
    memset(vis, 0, sizeof(vis));
    for (int i = 0; i < vec[s].size(); i++){
        int e = vec[s][i].e, w = vec[s][i].w;
        len[e] = w;
    }
    len[s] = 0; vis[s] = 1;
    
    for (int i = 0; i < n; i++){
        int min = INT_MAX, p = 1;
        for (int j = 1; j <= n; j++){
            if (len[j] < min&&vis[j] == 0){
                min = len[j]; p = j;
            }
        }

        vis[p] = 1;

        for (int j = 0; j < vec[p].size(); j++){
            int e = vec[p][j].e, w = vec[p][j].w;
            if (len[e]>len[p] + w&&vis[e] == 0){
                len[e] = len[p] + w;
            }
        }
    }
}
int main()
{
    
    while (cin >> n >> m, n+m){
        for (int i = 1; i <= n;i++)
            vec[i].clear();
        for (int i = 0; i < m; i++){
            int a, b, c;
            cin >> a >> b >> c;
            vec[a].push_back(Node(b, c));
            vec[b].push_back(Node(a, c));
        }
        dijkstra(1);
        cout << len[n] << endl;
    }
    return 0;
}

 

  • Floyd算法:
#include<cstdio>
#include<vector>
#include<cstring>
#include<climits>
#include<iostream>
using namespace std;
int main()
{
    int n, m;
    long map[105][105];
    while (cin >> n >> m, n+m){
        //memset(map, 125, sizeof(map));
        for (int i = 1; i <= n; i++){
            for (int j = 1; j <= n; j++)
                map[i][j] = INT_MAX;
            map[i][i] = 0;
        }
        for (int i = 0; i < m; i++){
            int a, b, c;
            cin >> a >> b >> c;
            map[a][b] = map[b][a] = c;
        }
        for (int k = 1; k <= n; k++)
        for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++){
            if (map[i][k] + map[k][j] < map[i][j])
                map[i][j] = map[i][k] + map[k][j];
        //    printf("%10d\t", map[i][j]);
        }
        cout << map[1][n] << endl;
    }
    return 0;
}

最后一个待调试~

hdu 2544 最短路

标签:需要   table   limit   包括   back   所在地   font   i++   sample   

原文地址:https://www.cnblogs.com/zengguoqiang/p/8858846.html

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